VOTING AND APPORTIONMENT

CHAPTER FIFTEEN

VOTING AND APPORTIONMENT Exercise Set 15.1 1. When a candidate receives more than 50% of the votes. 2. Each voter votes for one candidate. The candidate receiving the most votes is declared the winner. 3. Voters rank candidates from most favorable to least favorable. Each last place vote is awarded one point, each next to last place vote is awarded two points, each third from last place vote is awarded three points, etc. The candidate receiving the most points is the winner. 4. Each voter votes for one candidate. If a candidate receives a majority of votes, that candidate is declared the winner. If no candidate receives a majority, eliminate the candidate with the fewest votes. (If there is a tie for the fewest votes, eliminate all tied candidates.) Repeat this process until a candidate receives a majority. 5. Voters rank the candidates. A series of comparisons in which each candidate is compared to each of the other candidates follows. If candidate A is preferred to candidate B, then A receives one point. If candidate B is preferred to candidate A, then B receives one point. If the candidates tie, each receives ½ point. The candidate receiving the most points is declared the winner. 6. Different systems can lead to a different winner. 7. A preference table summarizes the results of an election. 8. a) Pair-wise comparison method 2 3 4 1 3 2 3 3+2+1 = 6 groupings 4 4 2 3 3 4 4 5 6 1 4 2 5 3 5 4 5 5+4+3+2+1 = 15 groupings 5 6 6 6 6 9. a) Jeter is the winner; he received the most votes using the plurality method.

b) No. 265128 265128

0.40192827 210361 265128 668316

= ≈+ +

is not a majority. Majority is > 334,158 votes.

10. a) Felicia is the winner. Felicia received the most votes using the plurality method.

b) No. 2863 2863

0.252192 2562 1671 2863 1959 11247

= ≈+ + + +

is not a majority. Majority is > 5,624 votes.

491

492 CHAPTER 15 Voting and Apportionment

11.

Number of votes 3 1 2 2 1

First B A C C A Second A B B A C Third C C A B B

12.

Number of votes 2 3 2 1

First A C B C Second B A A B Third C B C A

13. 9 + 5 + 3 + 2 = 19 employees 14. No. Mop had the most with 9 votes, but 9/19 = 0.47 which is not a majority. Majority is > 10. 15. Votes – (M): 9, (V): 5+3 = 8, (W): 2. Mop wins with the most votes.

16. M: 9 1st place votes = (9)(3) = 27 5 2nd place votes = (5)(2) = 10 5 3rd place votes = (5)(1) = 5 V: 8 1st place votes = (8)(3) = 24 11 2nd place votes = (11)(2) = 22 B: 2 1st place votes = (2)(3) = 6 3 2nd place votes = (3)(2) = 6 14 3rd place votes = (14)(1) = 14 M = 42 points; V = 46 points; B = 26 points Vacuum wins with 46 points.

17. A majority out of 19 votes is 10 or more votes. First choice votes: (M) 9, (V) 8, (B) 0 None receives a majority, thus B with the least votes is eliminated. Second round: (M) 9, (V) 5+3+2 = 10 Vacuum wins with a majority of 10 votes.

18. M vs. V: M = 9 V = 5+3+2 = 10 V gets 1 pt. M vs. B: M = 9+5 = 14 B = 3+2 = 5 M gets 1 pt. V vs. B: V = 9+5+3 = 17 B = 2 V gets 1 pt. Vacuum wins with 2 points.

19. B: 4 1st place votes = (4)(3) = 12 2 2nd place votes = (2)(2) = 4 3 3rd place votes = (3)(1) = 3 G: 2 1st place votes = (2)(3) = 6 4 2nd place votes = (4)(2) = 8 3 3rd place votes = (3)(1) = 3 M: 3 1st place votes = (3)(3) = 9 3 2nd place votes = (3)(2) = 6 3 3rd place votes = (3)(1) = 3 B = 19 points; G = 17 points; M = 18 points Beach wins with 19 points.

20. Votes – (B): 3+1 = 4, (G): 2, (M): 2+1 = 3 Beach wins with the most votes. 21. B vs. G: M = 3+2+1 = 6 G = 2+1 = 3 B gets 1 pt. B vs. M: B = 3+1 = 4 M = @+@+! = 5 M gets 1 pt. G vs. M: G = 3+2 = 5 M = 2+1+1 = 4 G gets 1 pt. All get 1 point, which indicates no winner.

SECTION 15.1 493

22. A majority out of 9 votes is 5 or more votes. First choice votes: (B) 4, (G) 2, (M) 3 None receives a majority, thus G with the least votes is eliminated. Second round: (B) 4, (M) 2+2+1 = 5 Mount Rushmore wins with a majority of 5 votes.

23. Votes: (S) 8+3+2 = 13 (L) 6+3 9 (H) 4+3+2 = 9 (T) 1 San Antonio wins with the most votes.

24. S: 9 1st place votes = (13)(4) = 52 5 2nd place votes = (5)(3) = 15 4 3rd place votes = (4)(2) = 8 10 4th place votes = (10)(1) = 10 L: 9 1st place votes = (9)(4) = 36 18 2nd place votes = (18)(3) = 54 4 3rd place votes = (4)(2) = 8 1 4th place vote = (1)(1) = 1 H: 9 1st place votes = (9)(4) = 36 9 2nd place votes = (9)(3) = 27 11 3rd place votes = (11)(2) = 22 3 4th place vote = (3)(1) = 3 T: 1 1st place votes = (1)(4) = 4 0 2nd place votes = 0 13 3rd place votes = (13)(2) = 26 18 4th place vote = (18)(1) = 18 S = 85 points; L = 99 points; H = 88 points; T = 48 points Los Angeles wins with 99 points.

25. A majority out of 32 votes is 16 or more votes. First choice votes: (S) 13, (L) 9, (H) 9, (T) 1 None receives a majority, thus T with the least votes is eliminated. Second round: (S) 13, (L) 9, (H) 10 No majority, thus eliminate L. Third round: (S) 16, (H) 16 Since S and H tied, there is no winner. 26. S vs. L: S = 8+3+2+1+2 = 16 L = 6+3+4+3 = 16 S / L get 0.5 pt. S vs. H: S = 8+3+3+2 = 16 H = 6+4+3+1+2 = 16 S / H get 0.5 pt. S and T: S = 8+3+2+1+2 = 16 T = 6+4+1 = 11 S gets 1 pt. L and H: L = 8+6+3+3 = 20 H = 4+3+2+1+2 = 12 L gets 1 pt. L and T: L = 8+6+3+4+3+3+2+2 = 31 T = 1 L gets 1 pt. H and T: H = 8+6+3+4+3+2+2 = 28 T = 4 H gets 1 pt. S = 2 H = 1.5 L = 2.5 T = 0 LA wins.

27. W: 5 1st place votes = (5)(3) = 15 4 2nd place votes = (4)(2) = 8 3 3rd place votes = (3)(1) = 3 D: 1 1st place votes = (1)(3) = 3 7 2nd place votes = (7)(2) = 14 4 3rd place votes = (4)(1) = 4 J: 6 1st place votes = (6)(3) = 18 1 2nd place votes = (1)(2) = 2 5 3rd place votes = (5)(1) = 5 W = 26 points; D = 21 points; J = 25 points Williams wins with 26 points.

28. Votes: (W): 5, (D): 1, (J): 4 + 2 = 6 Johnson wins with the most votes. 29. W vs. D: W = 5+4 = 9 D = 1+2 = 3 W gets 1 pt. W vs. J: W = 5 J = 1+4+2 = 7 J gets 1 pt. D vs. J: D = 5+1 = 6 J = 4+2 = 6 D and J get 0.5 pt. W = 1 pt. D = 1 pt. J = 1.5 pts. Johnson wins with 1.5 points.


30. A majority out of 12 votes is 6 or more votes. First choice votes: (W) 5, (D) 1, (J) 6 None receives a majority, thus D with the least votes is eliminated. Second round: (W) 5, (J) 1+4+2 = 7 Johnson wins with a majority of 7 votes.

31. A majority out of 12 votes is 6 or more votes. Most last place votes: (W) 3, (D) 4, (J) 5 Thus J with the most last place votes is eliminated. Second round using the most last place votes: (W) 1+2 = 3, (D) 5+4 = 9 Williams wins with the least last place votes.

32. Votes: (L): 5, (E): 2, (O): 4. Lehigh Road wins with the most votes. 34. A majority out of 11 votes is 6 or more votes. First choice votes: (L) 5, (E) 2, (O) 4 None receives a majority, thus E with the least votes is eliminated. Second round: (L) 5, (O) 2+4 = 6 Ontario Road wins with a majority of 6 votes.

33. L: 5 1st place votes = (5)(3) = 15 6 3rd place votes = (6)(1) = 6 E: 2 1st place votes = (2)(3) = 6 9 2nd place votes = (9)(2) = 18 O: 4 1st place votes = (4)(3) = 12 2 2nd place votes = (2)(2) = 4 5 3rd place votes = (5)(1) = 5 L = 21 points; E = 24 points; O = 21 points Erie Road wins with 24 points.

35. L vs. E: L = 5 E = 2+4 = 6 E gets 1 pt. L vs. O: L = 5 O = 2+4 = 6 O gets 1 pt. E vs. O: E = 5+2 = 6 O = 4 E gets 1 pt. Erie Road wins with 2 points.

36. A majority out of 11 votes is 6 or more votes. Most last place votes: (L) 2+4 = 6, (E) 0, (O) 5 Thus L with the most last place votes is eliminated. Second round using the most last place votes: (E) 0, (O) 4 Erie Road wins with the least last place votes.

37. a) Votes: (TI): 10, (C): 3, (HP): 2 Texas Instruments wins with the most votes. b) TI: 10 1st place votes = (10)(4) = 40 5 2nd place votes = (5)(3) = 15 C: 3 1st place votes = (3)(4) = 12 6 2nd place votes = (6)(3) = 18 6 3rd place votes = (6)(2) = 12 S: 9 3rd place votes = (9)(2) = 18 6 4th place votes = (6)(1) = 6 9 4th place votes = (9)(1) = 9 HP: 2 1st place votes = (2)(4) = 8 4 2nd place votes = (4)(3) = 12 9 4th place votes = (9)(1) = 9 TI= 55 points; C = 42 points; S = 24 points, HP = 29 points TI wins with 55 points.

37. c) A majority out of 15 votes is 8 or more votes. First choice votes: (TI) 10, (C) 3 (S) 0, (HP) = 2 Because TI already has a majority, TI wins. d) TI vs. C: TI = 6+4+2 = 12 C = 3 TI gets 1 pt. TI vs. S: TI = 6+4+3+2 = 15 TI gets 1 pt. TI vs. HP: TI = 6+4+3 = 14 HP = 2 TI gets 1 pt. C vs. S: C = 6+4+3+2 = 15 C gets 1 pt. C vs. HP: C = 6+3 = 9 HP = 4+3 = 7 C = gets 1 pt. S vs. HP: S = 6+3 = 9 HP = 4+2 = 6 S gets 1 pt. TI wins with 3 points.

SECTION 15.1 495

38. a) Votes: (L): 8, (M): 2, (S): 3, (H): 4 I Love Lucy wins with the most votes. b) L: 8 1st place votes = (8)(4) = 32 9 4th place votes = (9)(1) = 9 M: 2 1st place votes = (2)(4) = 8 15 2nd place votes = (15)(3) = 45 S: 3 1st place votes = (3)(4) = 12 2 2nd place votes = (2)(3) = 6 12 3rd place votes = (12)(2) = 24 H: 4 1st place votes = (4)(4) = 16 5 3rd place votes = (5)(2) = 10 8 4th place votes = (8)(1) = 8 L= 41 points; M = 53 points; S = 42 points, H = 34 points Mash wins with 53 points.

38. c) A majority out of 17 votes is 9 or more votes. First choice votes: (L) 8, (M) 2 (S) 3, (H) = 4 None receives a majority, thus M with the least votes is eliminated. Second round: (L) 8, (S) 5, (H) 4 No majority, thus eliminate H. Third round: (L) 8, (S) 9 Seinfeld wins with 9 votes. d) L vs. M: L = 8 M = 9 M gets 1 pt. L vs. S: L = 8 S = 9 S gets 1 pt. L vs. H: L = 8 H = 9 H gets 1 pt. M vs. S: M = 14 S = 3 M gets 1 pt. M vs. H: M = 13 H = 4 M gets 1 pt. S vs. H: S = 13 H = 4 S gets 1 pt. Mash wins with 3 points.

39. a) A: 6 1st place votes = (6)(4) = 24 1 2nd place vote = (1)(3) = 3 2 3rd place votes = (2)(2) = 4 5 4th place votes = (5)(1) = 5 B: 1 1st place vote = (1)(4) = 4 4 2nd place vote = (4)(3) = 12 9 3rd place votes = (9)(2) = 18 C: 5 1st place votes = (5)(4) = 20 6 2nd place vote = (6)(3) = 18 1 3rd place vote = (1)(2) = 2 2 4th place votes = (2)(1) = 2 D: 2 1st place votes = (2)(4) = 8 3 2nd place vote = (3)(3) = 9 2 3rd place votes = (2)(2) = 4 7 4th place votes = (7)(1) = 7 A = 36 points; B = 34 points; C = 42 points; D = 28 points C wins with 42 points.

39. b) Votes: (A): 6, (B): 1, (C): 5, (D): 2 A wins with the most votes. c) A majority out of 14 votes is 7 or more votes. First choice votes: (A) 6, (B) 1 (C) 5, (D) = 2 None receives a majority, thus B with the least votes is eliminated. Second round: (A) 7, (C) 5, (D) 2 No majority, thus eliminate D. Third round: (A) 9, (C) 5 A wins with 9 votes. d) A vs. B: A = 6 B = 8 B gets 1 pt. A vs. C: A = 9 C = 5 A gets 1 pt. A vs. D: A = 7 D = 7 A / D get 0.5 1 pt. B vs. C: B = 34 C = 11 C gets 1 pt. B vs. D: B = 9 D = 5 B gets 1 pt. C vs. D: C = 12 D = 2 C gets 1 pt. B and C tie with 2 points.

40. a) G vs. A: G = 69 A = 73 A gets 1 pt. G vs. C: G = 43 C = 99 C gets 1 pt. G vs. D: G = 43 D = 99 D gets 1 pt. A vs. C: A = 73 C = 69 A gets 1 pt. A vs. D: A = 73 D = 69 A gets 1 pt. C vs. D: C = 72 D = 70 C gets 1 pt. Apple wins with 3 points.

40. b) A majority out of 142 votes is 71 or more votes. First choice votes: G=43, A=30, C=29, D=40 None receives a majority, thus C with the least votes is eliminated. Second round: (G) 43, (C) 30, (D) 69 No majority, thus eliminate C. Third round: (G) 43, (C) 99 Compaq wins with 99 votes.


40. c) G: 43 1st place votes = (43)(4) = 172 1 2nd place vote = (1)(3) = 3 26 3rd place votes = (26(2) = 52 73 4th place votes = (73)(1) = 73 A: 30 1st place vote = (30)(4) = 120 43 2nd place vote = (43)(3) = 129 29 3rd place votes = (29)(2) = 58 26 4th place votes = (26)(1) = 26 C: 29 1st place votes = (29)(4) = 116 40 2nd place vote = (40)(3) = 120 73 3rd place vote = (73)(2) = 146 D: 40 1st place votes = (40)(4) = 160 59 2nd place vote = (59)(3) = 177 43 4th place votes = (43)(1) = 43 G = 300 points; A = 333 points; C = 382 points; D = 380 points Compaq wins with 380 points.

40. d) Votes: (G): 43, (A): 30, (C): 29, (D): 40 Gateway wins with the most votes. e) You must choose the voting method prior to the election. 41. a) If there were only two columns then only two of the candidates were the first choice of the voters. If each of the 15 voters cast a ballot, then one of the voters must have received a majority of votes because 15 cannot be split evenly. b) An odd number cannot be divided evenly so one of the two first choice candidates must receive more than half of the votes.

42. a) A: 1 + 2 + 3 + 1 = 8 B. 3 + 1 + 4 + 3 = 11 C: 4 + 4 + 1 + 1 = 10 D: 2 + 3 + 2 + 4 = 11 B and D tie with 11 points. b) A: 0 + 1 + 3 + 1 = 5 B: 3 + 0 + 5 + 3 = 11 C: 5 + 5 + 0 + 0 = 10 D: 1 + 3 + 1 + 5 = 10 B wins with 11 points. C wins with 42 points.

43. a) C: 4 + 1 +1 = 6 R: 4 + 4 + 3 = 11 W: 3 + 3 + 2 + 2 + 1 + 1 = 12 T: 4 + 3 + 2 + 2 = 11 The Warriors finished 1st, the Rams and the Tigers tied for 2nd , and the Comets were 4th. b) C: 5 + 0 = 5 R: 5 + 5 + 3 = 13 W: 3 + 3 + 1 + 1 + 0 + 0 = 8 T: 5 + 3 + 1 1 = 10 Rams - 1st, Tigers - 2nd, Warriors - 3rd, and Comets - 4th.

44. a) Each voter casts 3+2+1 = 6 votes. (20)(6) = 120 votes b) 120 – (55+25) = 120 – 80 = 40 votes c) No. Candidate B cannot win because the votes for Candidate A > votes for Candidate B.

45. a) Each voter casts $+3+2+1 = 10 votes. (15)(10) = 150 votes b) 150 – (35+40+25) = 150 – 100 = 50 votes c) Yes. Candidate D has more votes than each of the other 3 candidates.

46. A = 10 B = 7 C = 5 D = 9 Candidates A and D will win.

Exercise Set 15.2 1. If a candidate receives a majority of first place votes, then that candidate should be declared the winner. 2. A candidate who wins a first election and then gains additional support without losing any of the original support should also win a second election. 3. If a candidate is favored when compared individually with every other candidate, then that candidate should be declared the winner. 4. If a candidate is declared the winner of an election, and in a second election, one or more of the other candidates is removed, then the previous winner should still be declared the winner

SECTION 15.2 497

5. A candidate that is preferred to all others will win each pairwise comparison and be selected with the pairwise comparison method. 6. A candidate that holds a majority of first place votes wins each pairwise comparison and is selected with the pairwise comparison method. 7. If a candidate receives a majority of first place votes, then that candidate should be declared the winner. Plurality counts only the 1st place votes. 8. If a majority is not reached on the 1st vote, then the candidate with the lowest vote total is eliminated and successive votes are taken until one of the candidates achieves a majority vote. 9. The plurality method yields Tacos are the winner with a majority of 8 1st place votes. However, if the Borda count method is used: Tacos (8)(3) + (3)(2) + (4)(1) = 24 + 6 + 4 = 34 Pizza (4+3)(4) + (8)(2) = 28 + 16 = 44 Burgers (4)(2) + (8+3)(1) = 8 + 11 = 19 The winner is Pizza using the Borda count method, thus violating the majority criterion. 10. a) Total votes = 2+4+2+3 = 11 A vs. B: A = 4+2 = 6 B = 2+3 = 5 A gets 1 pt. A vs. C: A = 2+4 = 6 C = 2+3 = 5 A gets 1 pt. B vs. C: B = 2+4 = 6 C = 2+3 = 5 B gets 1 pt. Plan A wins with 2 points. b) C wins by a plurality of 5 votes. No, the head-to-head criterion is not satisfied.

11. Total votes = 3+2+1+1 = 7 Candidates A is the candidate of choice with a plurality of 4 votes. A: 4 1st place votes = (4)(4) = 16 3 4th place votes = (3)(1) = 3 B: 3 1st place vote = (3)(4) = 12 4 2nd place vote = (4)(3) = 12 C: 2 2nd place vote = (2)(3) = 6 4 3rd place vote = (4)(2) = 8 1 4th place vote = (1)(1) = 1 D: 1 2nd place vote = (1)(3) = 3 3 3rd place votes = (3)(2) = 6 3 4th place votes = (3)(1) = 3 G = 300 points; A = 333 points; A = 19 votes; B = 24 votes; C = 15 votes; D = 12 votes Candidate B is chosen with 24 votes, therefore the majority criterion is not satisfied.

12. a) Total votes = 12+6+4+3 = 25 B vs. W: B = 12+6+4 = 22 W = 3 B gets 1 pt. B vs. S: B= 12+3 = 15 S = 10 B gets 1 pt. B vs. R: B = 12+6 = 18 R = 7 B gets 1 pt. W vs. S: W = 12+3 = 15 S = 10 W gets 1 pt. W vs. R: W = 12+6+3 = 21 R = 4 W gets 1 pt. S vs. R: S = 12+6 = 18 S = 7 S gets 1 pt. Beach wins with 3 points. b) B wins by a plurality of 12 votes. Yes, the head-to-head criterion is satisfied.

13. P: 4 1st place votes = (4)(3) = 12 2 2nd place votes = (2)(2) = 4 3 3rd place votes = (3)(1) = 3 L: 3 1st place vote = (3)(3) = 9 5 2nd place vote = (5)(2) = 10 1 3rd place vote = (1)(1) = 1 S: 2 1st place votes = (2)(3) = 6 2 2nd place vote = (2)(3) = 6 5 3rd place vote = (5)(1) = 5 P = 19 votes; L = 20 votes; S = 17 votes P vs. L: P = 4+1 = 5 L = 4 P gets 1 pt. P vs. S: P = 4+1 = 5 S = 4 P gets 1 pt. L vs. S: L = 4+1+2 = 7 S = 2 L gets 1 pt. Because Parking wins by head-to-head comparison and the Lounge Areas win by Borda count method, the head-to-head criterion is not satisfied.


14. A: 2 1st place votes = (2)(3) = 6 72nd place votes = (7)(2) = 14 B: 2 1st place vote = (2)(3) = 6 2 2nd place vote = (2)(2) = 4 5 3rd place vote = (5)(1) = 5 C: 5 1st place votes = (5)(3) = 15 4 3rd place vote = (4)(1) = 4 A = 20 votes; B = 15 votes; C = 19 votes A vs. B: A = 2+2 = 4 B = 5 B gets 1 pt. A vs. C: A = 2+2 = 4 C = 5 C gets 1 p B vs. C: B = 2+2 = 4 C = 5 C gets 1 pt. Because C wins by head-to-head comparison and the A wins by the Borda count method, the head-to-head criterion is not satisfied.

15. A majority out of 25 votes is 13 or more votes. First choice votes: A=7, B=15, C=3 Since B has > 13 votes, B wins by plurality with elimination. A vs. B: A = 7+3 = 10 B = 15 B gets 1 pt. A vs. C: A = 7 C = 25-7 = 18 C gets 1 pt. B vs. C: B = 15+7 = 22 C = 3 B gets 1 pt. Yes, because B wins by both methods, the head-to-head criterion is satisfied.

16. A majority out of 25 votes is 13 or more votes. First choice votes: (A) 10, (B) 2, (C) 8, (D) = 5 None receives a majority, thus B with the least votes is eliminated. Second round: (A) 10, (C) 10, (D) 5 Still no majority, thus eliminate D. Third round: (A) 10, (C) 15 C wins with a majority of 15 votes. A vs. B: A = 10 B = 15 B gets 1 pt. A vs. C: A = 10 C = 15 C gets 1 pt. A vs. D: A = 12 D = 13 D gets 1 pt. B vs. C: B = 17 C = 8 B gets 1 pt. B vs. D: B = 20 D = 5 B gets 1 pt. C vs. D: C = 10 D = 15 D gets 1 pt. B wins with 3 points. Therefore, the head-to-head criterion is not satisfied.

17. Votes: A: 8, B: 4, C: 5; thus, A wins. If B drops out, we get the following: Votes: A: 8, C: 4 + 5 = 9, thus C would win. The irrelevant alternatives criterion is not satisfied. 18. Votes: A: 3, B: 4, C: 5; thus B wins If C drops out, we get the following: Votes: A: 3 + 5 = 8, B:6, thus A would win. The irrelevant alternatives criterion is not satisfied. 19. A receives 53 points, B receives 56 points, and C receives 53 points. Thus, B wins using the Borda count method. If A drops out, we get the following: B receives 37 points, and C receives 44 points. Thus, C wins the second vote. The irrelevant alternatives criterion is not satisfied.

20. A receives 38 points, B receives 35 points, C receives 35 points. Thus, A wins using the Borda count method. If B drops out we get the following: A receives 25 points, and C receives 29 points. Thus, C wins the second vote. The irrelevant alternatives criterion is not satisfied.

21. A majority out of 32 voters is 16 or more votes. Votes: A: 8 + 3 = 11, B: 9, C: 12; none has a majority, thus eliminate B. Votes: A:8 + 3 = 11, C: 9 +12 = 21, thus C wins. If the three voters who voted for A,C,B change to C,A,B, the new set of votes becomes: Votes: A: 12, B: 9, C: 11; none has a majority, thus eliminate B. Votes: A: 9 + 12 = 21, C = 11, thus A wins. Thus, the monotonicity criterion is not satisfied.

SECTION 15.2 499

22. A majority out of 29 voters is 15 or more votes. Votes: A: 8, B: 10, C: 11; none has a majority, thus eliminate A. Votes: B: 8 + 10 = 18, C: 7 + 4 = 11, thus B wins. After the four votes change their votes, the the new set of votes is A: 8, B: 14., C: 7; none has a majority, thus eliminate C. Votes: A: 7+ 8 = 15, B:14; thus A wins. Thus, the monotonicity criterion is not satisfied.

23. A majority out of 23 voters is 12 votes. Votes: A: 10, B: 8, C: 5; none has a majority, thus eliminate C. Votes: B: 10, B: 8 + 5 = 13; thus B wins. After A drops out, the new set of votes is B: 8, C: 10 + 5 = 15; thus C wins. The irrelevant alternatives criterion is not satisfied.

24. A majority out of 13 voters is 7 votes. Votes: A: 3, B: 6, C: 4; none has a majority, thus eliminate A. Votes: B: 6, C: 4 + 3 = 7; thus C wins. After B drops out, the new set of votes is Votes: A: 6 + 3 = 9, C: 4; thus A wins. The irrelevant alternatives criterion is not satisfied.

25. A receives 2 points, B receives 3 point, C receives 2 points, D receives 1 point, and E receives 2 pts. B wins by pairwise comparison. After A, C and E drop out, the new set of votes is B: 2 D: 3, thus D wins. The irrelevant

alternatives criterion is not satisfied. 26. A receives 3 points, B receives 1 point, C receives 3 points, D receives 1 point, and E receives 2 points. A and C tie, but when A vs. C, C wins and thus we declare C the winner. After A, B and E drop out, the new set of votes is table is C: 2 + 1 = 3, D: 4, thus D wins. The irrelevant alternatives criterion is not satisfied. 27. Total votes = 7 A wins with a majority of 4 votes. A: 4 1st place votes = (4)(3) = 12 3 3rd place votes = (3)(1) = 3 B: 2 1st place vote = (2)(3) = 6 5 2nd place vote = (5)(2) = 10 C: 1 1st place votes = (1)(3) = 3 2 2nd place votes = (2)(2) = 4 4 3rd place vote = (4)(1) = 4 A = 15 points; B = 16 points; C = 11 points B wins with 16 points. No. The majority criterion is not satisfied.

28. Total votes = 11 B wins with a plurality of 5 votes. A: 1 1st place votes = (1)(3) = 3 5 2nd place votes = (5)(2) = 10 5 3rd place votes = (5)(1) = 5 B: 6 1st place vote = (6)(3) = 18 5 3rd place votes = (5)(1) = 5 C: 4 1st place votes = (4)(3) = 12 6 2nd place votes = (6)(2) = 12 1 3rd place vote = (1)(1) = 1 A = 21 points; B = 23 points; C = 25 points C wins with 25 points. No. The majority criterion is not satisfied. 29. Total votes = 31 Majority = 16 or more a) Museum of Natural History b) Museum of Natural History c) Museum of Natural History d) None of them


30. a) Total votes = 44 A majority is > 22 votes. A: 8 1st place votes = (8)(5) = 40 8 3rd place votes = (8)(3) = 24 8 4th place votes = (8)(2) = 16 20 5th place votes = (20)(1) = 20 B: 20 1st place vote = (20)(5) = 100 2 2nd place vote = (2)(4) = 8 14 4th place votes = (14)(2) = 28 8 5th place votes = (8)(1) = 8 C: 4 1st place votes = (4)(5) = 20 8 2nd place votes = (8)(4) = 32 16 3rd place vote = (16)(3) = 48 8 4th place votes = (8)(2) = 16 8 5th place votes = (8)(1) = 8 D: 4 1st place votes = (4)(5) = 20 28 2nd place votes = (28)(4) = 112 4 3rd place votes = (4)(3) = 12 8 5th place votes = (8)(1) = 8 E: 8 1st place votes = (8)(5) = 40 2 2nd place votes = (2)(4) = 8 16 3rd place votes = (16)(3) = 48 14 4th place votes = (14)(2) = 28 A = 100 pts.; B = 136 pts.; C = 124 pts.; D = 152 pts.; E = 124 pts. Dow Chemical is chosen with 152 points. b) Burrows-Welcome will be chosen. c) Yes.

31. a) A majority out of 82 votes is 41 or more votes. First choice votes: (A) 28, (C) 30, (D) 24 None receives a majority, thus D with the least votes is eliminated. Second round: (A) 52, (C) 30 Thus, Jennifer Aniston is selected.. b) No majority on the 1st vote; C is eliminated with the fewest votes. Second round: (A) 38, (D) 44 Denzel Washington is chosen. c) Yes. 32. a) A receives 1 point, B receives 2½ points, C receives 1½ points, D receives 3 points, E receives 2 points. Thus, (D) wins. b) A receives 0 points, B receives 2½ points, D receives 2 points, E receives 1½ points. Thus, B wins. c) Yes. 33. A candidate who holds a plurality will only gain strength and hold and even larger lead if more favorable votes are added. 34. Answers will vary (AWV). 35. AWV 36. AWV 37. AWV

38. A majority out of 11 voters is 6 or more votes. a) Votes: A: 9, B: 2; thus A wins. b) Votes: A: 4 + 2 = 6, C: 5; Yes, A wins. c) The five voters who favor C should vote C, B, A instead of C, A, B.

39. AWV

Exercise Set 15.3 1. If we divide the total population by the number of items to be apportioned we obtain a number called the standard divisor. 2. The standard quota is found by dividing each group’s population by the standard divisor. 3. The standard quota rounded down to the nearest whole number. 4. The standard quota rounded up to the nearest whole number. 5. An apportionment should always be either the upper quota or the lower quota. 6. Hamilton’s method 7. Jefferson’s method, Webster’s method, Adams’s method 8. a) Jefferson’s method b) Adam’s method c) Webster’s method 9. a) Webster’s method b) Adam’s method c) Jefferson’s method 10. Jefferson’s method, Webster’s method, Adams’s method

SECTION 15.3 501

11. a) 7500000

50, 000150

= = standard divisor

b) and c) State A B C D Total

Population 1,222,000 2,730,000 857,000 2,693,000 7,500,000 Standard Quota 24.40 54.60 17.14 53.86

Lower Quota 24 54 17 53 148

Hamilton’s Apportionment 24 55 17 54 150 12. a) and b)

State A B C D Total Population 1,222,000 2,730,000 857,000 2,693,000 7,500,000

Modified Quota 24.65 55.15 17.31 54.40 Jefferson’s Apportionment 24 55 17 54 150

(round down) 13. a) and b)


Modified Quota 24.70 55.26 17.35 54.51

Jefferson’s Apportionment 24 55 17 54 150 (round down)

14. a) and b)

State A B C D Total

Population 1,222,000 2,730,000 857,000 2,693,000 7,500,000 Modified Quota 24.11 53.95 16.94 53.22

Adams’ Apportionment 25 54 17 54 150

(round up) 15. a) and b)

State A B C D Total

Population 1,222,000 2,730,000 857,000 2,693,000 7,500,000

Modified Quota 24.06 53.85 16.90 53.12 Adams’ Apportionment 25 54 17 54 150

(round up) 16. a) and b)


Standard Quota 24.40 54.60 17.14 53.86

Webster’s Apportionment 24 55 17 54 150 (standard rounding)


17. a) and b)

State A B C D Total

Population 1,222,000 2,730,000 857,000 2,693,000 7,500,000 Modified Quota 24.38 54.55 17.12 53.81

Webster’s Apportionment 24 55 17 54 150

18. a) Standard divisor = total 675

2725 25

= =

b) and c) Hotel A B C Total

Amount 306 214 155 675

Standard Quota 11.33 7.93 5.74

Hamilton’s Apportionment 11 8 6 25 19. a) and b)

Hotel Al Bob Charlie Total

Amount 350 530 470 1350

Modified Quota 8.05 12.18 10.84 Jefferson’s Apportionment 8 12 10 30

(rounded down) 20. a) and b)

Hotel Al Bob Charlie Total Amount 350 530 470 1350

Modified Quota 8.14 12.33 10.93

Jefferson’s Apportionment 8 12 10 30 21. a) and b)

Hotel Al Bob Charlie Total

Amount 350 530 470 1350 Modified Quota 7.45 11.28 10.00

Adam’s Apportionment 8 12 10 30

(rounded up) 22. a) and b)

Hotel Al Bob Charlie Total Amount 350 530 470 1350

Modified Quota 7.29 11.04 9.79 Adam’s Apportionment 8 12 10 30

(rounded up)

SECTION 15.3 503

23. a) and b)

Store Al Bob Charlie Total

Amount 350 530 470 1350 Standard Quota 7.78 11.78 10.44

Webster’s Apportionment 8 12 10 30

(standard rounding) 24. a) and b)

Store Al Bob Charlie Total Amount 350 530 470 1350

Modified Quota 7.61 11.52 10.22 Webster’s Apportionment 8 12 10 30

25. a) A standard divisor = total 540

1830 30

= =

b) Store A B C D Total

Population 75 97 140 228 540

Standard Quota 4.177 5.39 7.78 12.67 30 26.

Store A B C D Total Population 123 484 382 271 1260

Standard Quota 5.86 23.05 18.19 12.90 Lower Quota 5 23 18 12 58

Hamilton’s Apportionment 6 23 18 13 60 27. A divisor of 20.5 was used.

Store A B C D Total

Population 123 484 382 271 1260 Modified Quota 6.00 23.61 18.63 13.22


28. A divisor of 21.5 was used.

Store A B C D Total

Population 123 484 382 271 1260

Modified Quota 5.72 27.51 17.77 12.60 Adams’ Apportionment 6 23 18 13 60

(round up) 29.

Store A B C D Total Population 123 484 382 271 1260

Standard Quota 5.86 23.05 18.19 12.90




52250 250

= =

b) School LA Sci. Eng. Bus. Hum Total

Enrollment 1746 7095 2131 937 1091 13000 Standard Quota 33.58 136.44 40.98 18.02 20.98

31.

School LA Sci. Eng. Bus. Hum Total

Enrollment 1746 7095 2131 937 1091 13000

Standard Quota 33.58 136.44 40.98 18.02 20.98 Lower Quota 33 136 40 18 20 247

Hamilton’s Apportionment 34 136 41 18 21 250 32. A divisor of 51.5 was used.

School LA Sci. Eng. Bus. Hum Total Enrollment 1746 7095 2131 937 1091 13000

Modified Quota 33.90 137.77 41.38 18.19 21.18

Jefferson’s Apportionment 33 137 41 18 21 250 (round down)

33. A divisor of 52.5 was used.

School LA Sci. Eng. Bus. Hum Total

Enrollment 1746 7095 2131 937 1091 13000 Modified Quota 33.26 135.14 40.59 17.85 20.78

Adam’s Apportionment 34 136 41 18 21 250

(round up) 34.

School LA Sci. Eng. Bus. Hum Total Enrollment 1746 7095 2131 937 1091 13000

Standard Quota 33.58 136.44 40.98 18.02 20.98 Webster’s Apportionment 34 136 41 18 21 250

(standard rounding)

35. a) A standard divisor = total 13500

90150 150

= =

Dealership A B C D Total Annual Sales 4800 3608 2990 2102 13500

Standard Quota 53.33 40.09 33.22 23.36 150.00 36.


Standard Quota 53.33 40.09 33.22 23.36 150.00

Hamilton’s Apportionment 53 40 33 24 150

SECTION 15.3 505

37.

Dealership A B C D Total

Annual Sales 4800 3608 2990 2102 13500 Standard Quota 53.33 40.09 33.22 23.36 150.00

Jefferson’s Apportionment 54 40 33 23 150 38.

Dealership A B C D Total

Annual Sales 4800 3608 2990 2102 13500 Standard Quota 53.33 40.09 33.22 23.36 150.00

Adam’s Apportionment 53 40 33 24 150 39.


Standard Quota 53.33 40.09 33.22 23.36 150.00



14210 210

= =

b) Precinct A B C D E F Total Crimes 743 367 432 491 519 388 2940

Standard Quota 53.07 26.21 30.86 35.07 37.07 27.71 41.

Precinct A B C D E F Total Crimes 743 367 432 491 519 388 2940

Standard Quota 53.07 26.21 30.86 35.07 37.07 27.71

Lower Quota 53 26 30 35 37 27 208 Hamilton’s Apportionment 53 26 31 35 37 28 210

42. The divisor 3.8 as used.

Precinct A B C D E F Total

Crimes 743 367 432 491 519 388 2940 Modified Quota 53.84 26.59 31.30 35.58 37.61 28.12

Jefferson’s Apportionment 53 26 31 35 37 28 210

(round down) 43. The divisor 14.2 as used.

Precinct A B C D E F Total Crimes 743 367 432 491 519 388 2940

Modified Quota 52.32 22.85 30.42 34.58 36.55 27.32 Adam’s Apportionment 53 26 31 35 37 28 210

(round up)


44.

Precinct A B C D E F Total

Crimes 743 367 432 491 519 388 2940 Standard Quota 52.32 22.85 30.42 34.58 36.55 27.32

Webster’s Apportionment 53 26 31 35 37 28 210

(standard rounding)


12200 200

= =

b) Shift A B C D Total Room calls 751 980 503 166 2400

Standard Quota 62.58 81.67 41.92 13.83 46.

Shift A B C D Total

Room calls 751 980 503 166 2400

Standard Quota 62.58 81.67 41.92 13.83 Lower Quota 62 81 41 13 197

Hamilton’s Apportionment 62 82 42 14 200 47. The divisor 11.9 was used.

Shift A B C D Total Room calls 751 980 503 166 2400

Modified Quota 63.11 82.35 42.27 13.95


48. The divisor 12.1 was used.


Modified Quota 62.07 80.99 41.57 13.72

Adam’s Apportionment 63 81 42 14 200 (round up)

49. The divisor 12.02 was used.


Modified Quota 62.48 81.53 41.85 13.81

Webster’s Apportionment 62 82 42 14 200 (standard rounding)

SECTION 15.4 507

50. Standard divisor = 3615920

34437.33105

=

a) Hamilton’s Apportionment: 7, 2, 2, 2, 8, 14, 4, 5, 10, 10, 13, 2, 6, 2, 18 b) Jefferson’s Apportionment: 7, 1, 2, 2, 8, 14, 4, 5, 10, 10, 13, 2, 6, 2, 19 c) States that Benefited: Virginia States Disadvantaged: Delaware Exercise set 15.4 1. The Alabama paradox occurs when an increase in the total # of items results in a loss of items for a group. 2. The new-states paradox occurs when the addition of a new group changes the apportionment of another group. 3. The population paradox occurs when group A loses items to group B, although group A’s population grew at a higher rate than group B’s. 4. Yes, it can produce the Alabama paradox, population paradox, and new-states paradox. 5. Hamilton’s, Jefferson’s 6. Adam’s, Webster’s

7. New divisor = 900

17.6551

=

School A B C D E Total Standard Quota 11.90 9.35 9.07 9.92 10.76

Lower Quota 11 9 9 9 10 48

Hamilton’s Apportionment 12 9 9 10 11 51 No. No school suffers a loss so the Alabama paradox does not occur.

8. a) Standard divisor = 2592

18144

=

School A B C D Total

Population 739 277 618 958 2592

Standard Quota 41.06 15.38 34.33 53.22 Hamilton’s Apportionment 41 16 34 53 144

b) New divisor = 2592

17.88145

=

School A B C D Total Population 739 277 618 958 2592

Standard Quota 41.33 15.49 34.56 53.57

Hamilton’s Apportionment 41 16 34 53 144 Yes. School B loses a monitor while schools C and D each gain a monitor.


3030

=

State A B C Total Population 161 250 489 900

Standard Quota 5.37 8.33 16.30 Hamilton’s Apportionment 6 8 16 30


9. b) New divisor = 900

29.0331

=



Hamilton’s Apportionment 5 9 17 31 Yes, state A loses 1 seat and states B and C each gain 1 seat.


5000200

=

State A B C Total

Population 233,000 461,000 306,000 1,000,000 Standard Quota 46.60 92.20 61.20

Lower Quota 46 92 61 199

Hamilton’s Apportionment 47 92 61 200

10. b) New divisor = 1000000

4975.12201

=

State A B C Total Population 233,000 461,000 306,000 1,000,000

Standard Quota 46.83 92.66 61.51 Lower Quota 46 92 61 199

Hamilton’s Apportionment 47 93 61 201 No. None of the States lost a seat.


125200

=

City A B C Total

Population 8130 4030 12,840 25,000 Standard Quota 65.04 32.24 102.72



125.625200

=

City A B C Total New Population 8150 4030 12,945 25,125


No. None of the Cities loses a bonus.


3030

=

College A B C Total

Faculty 162 249 489 900 Standard Quota 5.40 8.30 16.30



SECTION 15.4 509


32.16730

=

College A B C Total Faculty 178 269 518 965


Lower Quota 5 8 16 29 Hamilton’s Apportionment 6 8 16 30

No. The opportionment is the same.


10054

=

Division A B C D E Total

Population 733 1538 933 1133 1063 5400 Standard Quota 7.33 15.38 9.33 11.33 10.63

Lower Quota 7 15 9 11 10 52 Hamilton’s Apportionment 7 16 9 11 11 54


10154

=

Division A B C D E Total Population 733 1539 933 1133 1116

Standard Quota 7.26 15.238 9.238 11.22 11.05

Lower Quota 7 15 9 11 11 53 Hamilton’s Apportionment 8 15 9 11 11 54

Yes. Division B loses an internship Division A even though the population of division B grew faster than the population of division A.


120250

=



b) Same divisor = 30000

120250

=

State A B C Total

Population 464 10551 19100 30110 Standard Quota 3.87 87.93 159.17

Hamilton’s Apportionment 3 88 159 250 No. The opportionment is the same.



10048

=

Tech. Data A B Total Employees 844 3956 4800

Standard Quota 8.44 39.56

Lower Quota 8 39 47 Hamilton’s Apportionment 8 40 48


100.4455

=

Tech. Data A B C Total Employees 844 3956 724 5524



Yes. Group B loses a manager.


100100

=

State A B Total

Population 1135 8865 10000 Standard Quota 11.35 88.65

Hamilton’s Apportionment 11 89 100


100.24106

=



Yes. State C loses a seat to State B.


15, 00066

=

State A B C Total

Population 68970 253770 667260 990000



15,152.1171

=

State A B C D Total Population 68970 253770 667260 85800 1075800

Standard Quota 4.55 16.75 44.04 5.66

Hamilton’s Apportionment 4 17 44 6 71 Yes. State C loses a seat to State B.

REVIEW EXERCISES 511


10033

=

State A B Total Population 744 2556 3300 Standard Quota 7.44 25.56 Lower Quota 7 25 32 Hamilton’s Apportionment 7 26 33


100.2540

=

State A B C Total

Population 744 2556 710 4010 Standard Quota 7.42 25.50 7.08 Lower Quota 7 25 7 39 Hamilton’s Apportionment 7 26 7 40

No. The apportionment is the same. Review Exercises 1. a) Robert Rivera wins with the most votes (12). b) A majority out of 24 voters is 13 or more votes. Robert Rivera does not have a majority. 2. a) Michelle MacDougal wins with the most votes (224). b) Yes. A majority out of 421 voters is 211 or more votes. 3.

# of votes 3 2 1 3 1

First B A D C D Second A C C B A Third C D A A B Fourth D B B D C

4. # of votes 2 2 2 1

First C A B C Second A B C B Third B C A A

5. Number of votes = 6 + 4 + 3 +2 + 1 + 1 = 17 6. Park City wins with a plurality of 6 votes. 7. P: 50 points, V: 47 points, S: 35 points, A: 38 points. Park City wins with 50 points.

8. A majority out 17 voters is 9 or more votes. Votes: P: 6+1 = 7, V: 4, S: 3+2 = 5, A:1. None has a majority, thus eliminate A. Votes: P: 6+1 = 7, V: 4, S: 3+2+1 = 6 None has a majority, thus eliminate V. Votes: P: 6+4+1 = 11, S: 3+2+1 = 6. Park City wins.

9. P: 3 pts., V: 2 pts., S: 0 pts., A: 1 pt. Park City wins with 3 points.

10. Votes: P: 7, V: 4, S: 5, A: 1 None has a majority, thus eliminate S with most last place votes. Votes: P: 10, V: 4, A: 3; Park City wins.

11. 38+30+25+7+10 = 110 students voted 12. Volleyball wins with a plurality of 40 votes.

13. S: 223 pts., V: 215 pts., B: 222 pts. Soccer wins.

14. A majority out of 110 voters is 56 or more votes. Votes: S: 38, V: 40, B: 32; None has a majority, thus eliminate B. Votes: S: 45, V: 65 Volleyball wins.


15. S: 1 pt., V: 1 pt., B: 1 pt. A 3-way tie

16. Votes: S: 38, V: 40, B: 32 None has a majority, thus eliminate V with the most last place votes. Votes: S: 68, B: 42. Soccer wins.

17. a) Votes: A: 161+134 = 295, F: 45, M: 12, P: 0 AARP wins. b) Yes. A majority out of 372 voters is 186 or more votes. AARP receives a majority. c) A: 985 pts., F: 740 pts., M: 741 pts., P: 852 pts. AARP wins. d) 186 or more votes is needed for a majority. Votes: A: 295, F: 45, M: 12, P: 0 AARP wins. e) A: 3 pts., F: 1 pt., M: 1 pt., P: 1 pt. AARP wins.

18. Votes: (NO): 70, (LV): 55, (C): 30, (SD): 45 a) A majority out of 200 voters is 101 or more votes. None of the cities has a majority. b) New Orleans win a plurality of 70 votes. c) (NO):410 pts., (LV): 580 pts., (C): 505 pts., (SD): 495 pts. Las Vegas wins. d) Las Vegas wins with 130 pts. to 70 pts. for NO. e) NO: 0 pts., LV: 3 pts., C: 1 pt., SD: 1 pt. Las Vegas wins with points.

19. a) A majority out of 16 voters is 9 or more votes. Votes: (EB): 4+3+ = 7, (FW): 1+1 = 2, (G): 0, (WB): 6+1 = 7 None has a majority, thus eliminate G. Votes: (EB): 4+3 = 7, (FW): 1+1 = 2, (WB): 6 + 1 = 7 None has a majority, thus eliminate FW Votes: (EB): 4+3+1 = 8, (WB): 6+1+1 = 8. Thus, EB and WB tie. b) Use the Borda count method to break the tie. (EB) = 46 points, (WB) = 50 points; World Book wins.

19. c) (EB) vs. (WB): EB: 4+3+1 = 8 points, (WB): 6+1+1 = 8 points. EB and WB tie again. 20. A: 33 pts., B: 39 pts, C: 28 pts., D: 20 pts. Using the Borda count, method B wins. However, B only has 3 first place votes, thus the majority criterion is not satisfied. 21. In a head-to-head comparison, B must win over all the others. For (B vs. A), A wins with 3 pts. The head-to-head criterion is not satisfied.

22. a) A majority out of 42 voters is 21 or more votes. Votes: A: 12, B: 10+6 = 16, C: 14 None has the majority, thus eliminate A. Votes: B :10+6 = 16, C: 14+12 = 26 C wins. b) The new preference table is

Number of votes 10 14 6 12 First B C C A

Second A B B C Third C A A B

Votes: A: 12, B: 10, C: 20; None has a majority, thus eliminate B. Votes: A: 22, C: 20 A wins. When the order is changed A wins. Therefore, the monotonicity criterion is not satisfied.

22. c) If B drops out the new table is Number of votes 10 14 6 12

First A C C A Second C A A C

Votes: A: 10+12 = 22, C: 14+6 = 20 A wins. Since C won the first election and then after B dropped out A won, the irrelevant criterion is not satisfied.

REVIEW EXERCISES 513

23. a) M has 0 pts., S has 3 pts., F has 1 pt., and E has 1 pt. Thus, Starbucks wins. b) Maxwell House wins w/a plurality of 33 votes. c) M = 228 pts., S = 277 pts., F = 293 pts., and E = 292 pts. Thus, Folgers wins. d) Eight O’clock wins over Maxwell House with 76 points. e) Same results as in a), thus, Starbucks wins. f) The plurality, plurality with elimination, and Borda count methods all violate the head-to-head criterion. 25. The Borda count method 26. Plurality and plurality w/elimination methods 27. Pairwise comparison and Borda count methods

24. a) Yes. Fleetwood Mac is favored when compared to each of the other bands. b) Votes: A: 15, B: 34, C: 9+4 = 13, F: 25 Boston wins. c) A: 217 points, B: 198 points, C: 206 points, F: 249 points Fleetwood Mac wins. d) A majority out of 87 voters is 44 or more votes. Votes: A: 15, B: 34, C: 13, F:25 None has a majority, thus eliminate C. Votes: A: 15+9+4 = 28, B: 34, F: 25 None has a majority, thus eliminate F. Votes: A: 28+25 = 53, B: 34 Abba wins. e) A = 2 pts., B = 0 pts., C = 1 pt., F = 3 pts. Thus, Fleetwood Mac wins. f) Plurality and plurality w/elimination methods


60010

=

Region A B C Total Number of Houses 2592 1428 1980 6000



29. Using the modified divisor 500.

Region A B C Total

Number of Houses 2592 1428 1980 6000


(rounded down) 30. Using the modified divisor 700.

Region A B C Total

Number of Houses 2592 1428 1980 6000

Modified Quota 3.70 2.04 2.83 Adam’s Apportionment 4 3 3 10

(rounded up) 31. Using the modified divisor 575.

Region A B C Total Number of Houses 2592 1428 1980 6000


Webster’s Apportionment 5 2 3 10 (normal rounding)


32. Yes. Hamilton’s Apportionment becomes 5, 2, 4. Region B loses one truck.


3023

=

Course A B C Total

Number of Students 311 219 160 690 Standard Quota 10.37 7.30 5.33


Hamilton’s Apportionment 11 7 5 23 34. Use the modified divisor 28

Course A B C Total Number of Students 311 219 160 690


(round down) 35. Use the modified divisor 31.5

Course A B C Total Number of Students 311 219 160 690


Adam’s Apportionment 10 7 6 23 (round up)

36. Use the modified divisor 29.5

Course A B C Total

Number of Students 311 219 160 690 Modified Quota 10.54 7.42 5.42

Webster’s Apportionment 11 7 5 23

(standard rounding)

37. The new divisor is 698

30.3523

=

Course A B C Total

Number of Students 317 219 162 698 Standard Quota 10.44 7.22 5.34


Hamilton’s Apportionment 11 7 5 23 No. The apportionment remains the same.

38. The Standard divisor = 55000

100055

=

State A B Total Population 4862 50138 55,000

Standard Quota 4.86 50.14 Hamilton’s Apportionment 5 50 55

CHAPTER TEST 515

39. The apportionment is 4, 51. 40. The apportionment is 5, 50. 41. The apportionment is 5, 50.

42. The new divisor is 60940

1015.6760

=



Yes. State A. gains a seat while State B loses a seat. Chapter Test 1. 6+5+5+4 = 20 members voted. 2. No candidate has a majority of > 10 votes. 3. Chris wins with a plurality of 9 votes. 4. D = 41 pts., C = 44 pts., S = 35 pts. Chris wins. 5. Donyall wins with 11 pts. 6. D = 1.5 pts., C = 1 pt., S = 0.5 pt. Donyall wins. 7. a) Votes: H: 26+14 = 40, I: 29, L: 30, S: 43 Thus, the snail wins. b) (H) 1st (40)(4) = 160 2nd (59)(3) = 177 3rd (0)(2) = 0 4th (43)(1) = 43 H receives 380 points. ( I ) 1st (29)(4) = 116 2nd (40)(3) = 120 3rd (73)(2) = 146 4th (0)(1) = 0 I receives 382 points (L) 1st (30)(4) = 120 2nd (43)(3) = 129 3rd (43)(2) = 86 4th (26)(1) = 26 L receives 361 points

7. b) (S) 1st (43)(4) = 172 2nd (0)(3) = 0 3rd (26)(2) = 52 4th (73)(1) = 73 S receives 297 points. The iguana (I) wins with the most points. c) A majority out of 142 voters is 72 or more votes. Votes: H: 40, I: 29, L: 30, S: 43; None has a majority, thus eliminate I. Votes: H: 69, L: 30, S: 43 None has a majority, thus eliminate L. Votes: H: 99, S: 43 The hamster wins. d) H vs. I: I gets 1 pt. H vs. L: L gets 1 pt. H vs. S: H gets 1 pt. I vs. L: L gets 1 pt. I vs. S: I gets 1 pt. L vs. S: L gets 1 pt. Ladybug wins with 3 points.

8. Plurality: Votes: W: 86, X: 52+28 = 80, Y: 60, Z: 58 W wins. Borda count: W gets 594 points, X gets 760 points, Y gets 722 points, Z gets 764 points Z wins Plurality with elimination: A majority out of 284 voters is 143 or more votes. Votes: W: 86, X: 80, Y: 60, Z: 58 None has a majority, thus eliminate Z. Votes: W: 86, X: 80+58 = 138, Y: 60 None has a majority, thus eliminate Y. Votes: W: 86, X: 138+60 = 198 X wins.

8. Head-to-Head: When Y is compared to each of the others, Y is favored. Thus Y wins the head-to-head comparison. Plurality, Borda count and Plurality with elimination each violate the head-to-head criterion. The pairwise method never violates the head-to-head criterion. 9. A majority out of 35 voters is 18 or more votes. Louisiana (L) has a majority. However, Mississippi (M) wins using the Borda count method. Thus the majority criterion is violated.


10. a) The standard divisor = 33000

110030

=

State A B C Total Population 6933 9533 16534 33,000


Hamilton’s Apportionment 6 9 15 30 b)

State A B C Total

Population 6933 9533 16534 33,000 Modified Quota 6.30 8.67 15.03

Jefferson’s Apportionment 6 8 15 29

(round down) c) The new divisor 1064.52

State A B C Total

Population 6933 9533 16534 33,000 Standard Quota 6.51 8.96 15.53

Hamilton’s Apportionment 6 9 16 31 The Alabama paradox does not occur, sine none of the states loses a seat.

d) The divisor = 33826

1091.1631

=

State A B C Total Population 7072 9724 17030 33,826


Hamilton’s Apportionment 6 9 16 31 The Alabama paradox does not occur, sine none of the states loses a seat.

10. e) The new divisor is 38100

1058.3336

=

State A B C D Total Population 6933 9533 16534 5100 38100

Standard Quota 6.55 9.01 15.62 4.82 Hamilton’s Apportionment 6 9 16 5 36

The new states paradox does not occur, sine none of the existing states loses a seat.

VOTING AND APPORTIONMENT

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